Third law of thermodynamic as function of equilibrium constant

[kthouz]

Hello!
Here is an equation that i sometimes use in calculating enthalpy of reaction aA+bB=cC+dD.
a,b,c and d are stoechiometric coefficients, A,B,C and D are chemical compounds in reaction. They say that is an other form of the third law of thermodynamic, can one guide me please to derive it?
$$\Delta$$H=T[$$\Delta$$$$\Phi$$-Rln(K)].
where T is temperature, $$\Delta$$$$\Phi$$ is the Gibbs Free energy, R is the ideal gas constant and K is the partial pressure equilibrium constant.

 

[Aaron Barnard]

This equation is derived from the Third Law of Thermodynamics. According to the Third Law, the enthalpy change of a reaction is equal to the change in Gibbs free energy of the reaction at zero absolute temperature. Therefore, we can write:

\DeltaH = \DeltaG|_{T=0}

At non-zero temperatures, the enthalpy change of a reaction is equal to the change in Gibbs free energy plus an additional term that accounts for the entropy change of the reaction. This additional term is known as the "enthalpy of reaction" and is given by the equation:

\DeltaH = \DeltaG + T\DeltaS

Now, the entropy change of a reaction is related to the equilibrium constant of the reaction, K, through the equation:

\DeltaS = -Rln(K)

where R is the ideal gas constant. Therefore, combining the above equations, we can obtain the following equation for the enthalpy change of a reaction:

\DeltaH = \DeltaG + T[-Rln(K)]

which is equivalent to the equation you have provided:

\DeltaH=T[\Delta\Phi-Rln(K)].